Question

The dimensional formula for electric flux is
A. $M{L}^3{I}^{-1}{T}^{-3}$
B. $M^2{L}^2{I}^{-1}{T}^{-2}$
C. $M{L}^3{I}^1{T}^{-3}$
D. $M{L}^{-3}{I}^{-1}{T}^{-3}$

Answer 1

Hint: The dimensional formula is defined as the expression for the unit of a physical quantity in terms of fundamental quantities that are mass ($M$), length ($L$), and time ($T$). Here, we will use the formula of force to find the value of the electric field. Then, we will use the formula of electric flux to calculate the dimensional formula of the electric flux.

Formula used:
The formula for force is given by
$F=qE$
Where $F$ is the force, $q$ is the charge and $E$ is the electric field.
Also, the formula for the electric flux is given by
${\varphi }_E=E.S$
Here, ${\varphi }_E$ is the electric flux, $E$ is the electric field and $S$ is the surface area.

Complete step by step answer:
Now, the electric flux can be defined as the measure of the electric field that passes through a given surface area. Also, the electric flux can be defined product of the electric field and the area of the surface, which is shown below
Electric flux ${\varphi }_E=E.S$
$\Rightarrow {\varphi }_E=ES\cos \theta$
Here, $E$ represents the magnitude of the electric field, $S$ represents the area of the surface, and $\theta$ represents the angle between the electric field vector $\overrightarrow{E}$ and the area vector $d\overrightarrow{S}$ .
Now, the formula of the force is given by
$F=qE$
Therefore, we can calculate the electric field from the above equation as
$E={\displaystyle \frac{F}{q}}$
Therefore, putting this value in the electric flux, we get
Electric flux $={\displaystyle \frac{F}{q}}.S$
Now, as we know that the S.I. unit of the electric field is ${\displaystyle \frac{newton\times meter{r}^2}{coulomb}}={\displaystyle \frac{N{m}^2}{ch\arg e\times time}}$
Therefore the dimensional formula can be calculated as shown below
${\displaystyle \frac{newton\times meter{r}^2}{ch\arg e\times time}}={\displaystyle \frac{kg\times m{s}^{-2}\times m^2}{I\times \mathrm{s}}}=kg{m}^3{\mathrm{s}}^{-3}{I}^{-1}$
Now, $kg$ is the unit of mass, $meter$ is the unit of length, $\sec ond$ is the unit of time, and $ampere$ is the unit of the current. Therefore, the dimensional formula of the electric flux density is $M{L}^3{T}^{-3}{I}^{-1}$.

So, the correct answer is “Option D”.

Note:
An alternate method to calculate the dimensional formula of electric flux is by using Gauss’s law.
The electric flux according to Gauss’s law through a closed surface is shown below
$\varphi ={\displaystyle \frac{q_{closed}}{{\epsilon }_0}}$
Now, the dimensional formula for the charge, $q_{closed}=\left[IT\right]$
Also, the dimensional formula for the permittivity, ${\epsilon }_0=\left[M^{-1}{L}^{-3}{I}^2{T}^4\right]$
Therefore, the dimensional formula of electric flux is given by putting the dimensional formula of both the quantities as shown below
$\varphi ={\displaystyle \frac{IT}{M^{-1}{L}^{-3}{I}^2{T}^4}}=M{L}^3{I}^{-1}{T}^{-3}$
Which is the required dimensional formula.